The gauge pressure in your car tires is 2.70 ✕ 105 n/m2 at a temperature of 35.0°c when you drive it onto a ship in los angeles to be sent to alaska. what is their gauge pressure (in atm) later, when their temperature has dropped to −42.0°c? assume the tires have not gained or lost any air.

The final gauge pressure will be 1.74 atm

Explanation:

Assume air tire as an ideal gas, therefore, it is considered:

The expression to describe the ideal gas process is:

Where v is the specific volume or the inverse of the density p:

(2)

Then the P and T are the absolute pressure and temperature respectively. Rg represents the particular gas constant for air, Rg is equal to 287 J/kg-K. Take into account that P and T must be expressed in Pascal and Kelvin respectively.

By reorganizing  the expression (2) as below is doing:

Can be noticed that the product Rg*p is constant therefore relation P/T will also be constant. For two different states 1 and 2 of the ideal gas, it follows:

(3)

Here, states 1 and 2 will represent the state before and after arriving in Alaska.

Note that the temperatures given are in °C so it must be converted:

Note also that the pressure given is the gauge pressure therefore it must be expressed as absolute pressure:

Where Patm is the atmosphere pressure and is equal to 101325 Pa then for Pg of 2.7 10^5 Pa:

Solving equation (3) for pressure at state 2:

For the temperatures and pressure calculated values:

As is required the gauge pressure and not the absolute pressure:

Finally re expressing the pressure in atm units:

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr

snell's law

refractve index = sin incidence / sin refraction